题目
Write a class RecentCounter
to count recent requests.
It has only one method: ping(int t)
, where t represents some time in milliseconds.
Return the number of ping
s that have been made from 3000 milliseconds ago until now.
Any ping with time in [t - 3000, t]
will count, including the current ping.
It is guaranteed that every call to ping
uses a strictly larger value of t
than before.
Example 1:
Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]
Note:
- Each test case will have at most
10000
calls to ping
.
- Each test case will call
ping
with strictly increasing values of t
.
- Each call to ping will have
1 <= t <= 10^9
.
题目大意
写一个 RecentCounter 类来计算最近的请求。它只有一个方法:ping(int t),其中 t 代表以毫秒为单位的某个时间。返回从 3000 毫秒前到现在的 ping 数。任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内,包括当前(指 t 时刻)的 ping。保证每次对 ping 的调用都使用比之前更大的 t 值。
提示:
- 每个测试用例最多调用 10000 次 ping。
- 每个测试用例会使用严格递增的 t 值来调用 ping。
- 每次调用 ping 都有 1 <= t <= 10^9。
解题思路
- 要求设计一个类,可以用
ping(t)
的方法,计算 [t-3000, t] 区间内的 ping 数。t 是毫秒。
- 这一题比较简单,
ping()
方法用二分搜索即可。
代码
type RecentCounter struct {
list []int
}
func Constructor933() RecentCounter {
return RecentCounter{
list: []int{},
}
}
func (this *RecentCounter) Ping(t int) int {
this.list = append(this.list, t)
index := sort.Search(len(this.list), func(i int) bool { return this.list[i] >= t-3000 })
if index < 0 {
index = -index - 1
}
return len(this.list) - index
}
/**
* Your RecentCounter object will be instantiated and called as such:
* obj := Constructor();
* param_1 := obj.Ping(t);
*/